Total number of possible outcomes = 36
Total number of outcomes with sum even = 18
P(even) = 18/36 = 1/2
Answer: 1/2
Answer:
2^2 your welcome
Step-by-step explanation:
Answer:
∠ B = 85°
Step-by-step explanation:
Complementary angles sum to 90° , then
x - 23 + 2x + 29 = 90 , that is
3x + 6 = 90 ( subtract 6 from both sides )
3x = 84 ( divide both sides by 3 )
x = 28
Then
∠ B = 2x + 29 = 2(28) + 29 = 56 + 29 = 85°
Answer:
Proportion of games an NFL team wins.
Proportion of defective products in a shipment.
Proportion of voters of a certain candidate in an election.
Step-by-step explanation:
Proportion is a relationship between measures.
Examples in real life are:
Proportion of games an NFL team wins.
For example, this past regular season, the Buffalo Bills won 13 games out of 16, so the proportion of games won by the Bills is 13/16 = 0.8125.
Other possible examples are:
Proportion of defective products in a shipment.
Proportion of voters of a certain candidate in an election.
...
The question is incomplete, here is the complete question:
The half-life of a certain radioactive substance is 46 days. There are 12.6 g present initially.
When will there be less than 1 g remaining?
<u>Answer:</u> The time required for a radioactive substance to remain less than 1 gram is 168.27 days.
<u>Step-by-step explanation:</u>
All radioactive decay processes follow first order reaction.
To calculate the rate constant by given half life of the reaction, we use the equation:
where,
= half life period of the reaction = 46 days
k = rate constant = ?
Putting values in above equation, we get:
The formula used to calculate the time period for a first order reaction follows:
where,
k = rate constant =
t = time period = ? days
a = initial concentration of the reactant = 12.6 g
a - x = concentration of reactant left after time 't' = 1 g
Putting values in above equation, we get:
Hence, the time required for a radioactive substance to remain less than 1 gram is 168.27 days.